Question 3.1: Use the Laplace transform method to solve Example 2.3, i.e. ......

The Laplace transform of the displacement response z(t) is given by Eq. (3.13),

L\left[z(t)\right] = L\left[F(t)\right] \times \left(\frac{\underline{z} }{\underline{F} } \right)             (A)

where in this case, L[F(t)], the Laplace transform of the force, F(t), is that of a unit step function, which from the table is 1/s, multiplied by the magnitude of the step, P, so

L\left[z(t)\right] = \frac{P}{s}            (B)

From Eq. (3.4), the transfer function is

\left(\frac{\underline{z} }{\underline{F} } \right) = \frac{1}{m}\cdot \frac{1}{(s^{2}+2\gamma \omega _{n}s+\omega ^{2}_{n})}              (C)

Substituting Eqs (B) and (C) into Eq. (A),

L\left[z(t)\right] =\frac{P}{s} \times \frac{1}{m}\cdot \frac{1}{(s^{2}+2\gamma \omega _{n}s+\omega ^{2}_{n})} =\frac{P}{m}\left[ \frac{1}{s}\cdot \frac{1}{(s^{2}+2\gamma \omega _{n}s+\omega ^{2}_{n})} \right]                 (D)

Ignoring, for the present, the constant P/m, which simply multiplies the final answer,
we now require the inverse transform of

This will not usually appear in standard lists of transforms, and it must be written in the form of partial  fractions. The rules for forming partial fractions can be found in standard texts, often under ‘integration’, since this is a common method for breaking expressions down into smaller components before integration. In this case the rules say that the required partial fractions are given by:

\left[ \frac{1}{s}\cdot \frac{1}{(s^{2}+2\gamma \omega _{n}s+\omega ^{2}_{n})} \right] \equiv \frac{A}{s}+ \frac{Bs + C}{(s^{2}+2\gamma \omega _{n}s+\omega ^{2}_{n})}                 (E)

where A, B and C are constants to be found. Equation (E) is an identity, valid for any value of s. Writing it in the form

1 \equiv A (s^{2}+2\gamma \omega _{n}s+\omega ^{2}_{n}) + Bs^{2} + Cs             (F)

the constants, A, B and C can now be found, either by sequentially substituting any three values for s, say the numbers 1, 2 and 3, or by equating the constants, the coefficients of s and those of s², on each side of the equation, in turn. Using the latter method:

Equating constants: 1  =  A\omega ^{2}_{n}      or       A  =  \frac{1}{\omega ^{2}_{n}}

Equating coeffs of s: 0  =  2A\gamma \omega _{n} + C    or      C  = -\frac{1}{\omega ^{2}_{n}} (2\gamma \omega _{n})

Equating coeffs of s²: 0  =  A +B         so       B = -\frac{1}{\omega ^{2}_{n}}.

Substituting for A, B and C into the right side of Eq. (E) we have

\frac{1}{\omega ^{2}_{n}}\left[\frac{1}{s}- \frac{s+2\gamma \omega _{n}}{(s^{2}+2\gamma \omega _{n}s + \omega ^{2}_{n})} \right]          (G)

We now require the inverse transforms of the two terms in the square brackets. The first, 1/s, is no problem, but we cannot find the inverse transform of the second term in Appendix A and this requires a little more algebra. Writing (s^{2}+2\gamma \omega _{n}s + \omega ^{2}_{n}) as (s+\gamma \omega _{n})^{2} + \omega ^{2}_{d} where \omega ^{2}_{d} = \omega ^{2}_{n} (1- \gamma ^{2}), the expression (G) can be written as:

\frac{1}{\omega ^{2}_{n}}\left[\frac{1}{s}-\frac{s+\gamma \omega _{n}}{(s+\gamma \omega _{n})^{2}+ \omega ^{2}_{d}} -\frac{\left[\frac{\gamma }{\sqrt{1- \gamma ^{2}} } \right] \omega _{d} }{(s+\gamma \omega _{n})^{2}+ \omega ^{2}_{d}} \right]              (H)

The inverse transforms of the three parts of expression (H) can now be found, using Appendix A, and the result (now re-inserting the constant, P/m) is

z = \frac{P}{m\omega ^{2}_{n}}\left[1-e^{-\gamma \omega _{n}t}\left(\cos \omega _{d}t + \frac{\gamma }{\sqrt{(1- \gamma ^{2})} } \sin \omega _{d}t \right) \right]         (I)

agreeing with Eq. (G) of Example 2.3, since m \omega ^{2}_{n}= k.